Is ${661999}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {661999}= &&{6}\cdot100000+ \\&&{6}\cdot10000+ \\&&{1}\cdot1000+ \\&&{9}\cdot100+ \\&&{9}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {661999}= &&{6}(99999+1)+ \\&&{6}(9999+1)+ \\&&{1}(999+1)+ \\&&{9}(99+1)+ \\&&{9}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {661999}= &&\gray{6\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {6}+{6}+{1}+{9}+{9}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${661999}$ is divisible by $3$ if ${ 6}+{6}+{1}+{9}+{9}+{9}$ is divisible by $3$ Add the digits of ${661999}$ $ {6}+{6}+{1}+{9}+{9}+{9} = {40} $ If ${40}$ is divisible by $3$ , then ${661999}$ must also be divisible by $3$ Add the digits of ${40}$ $ {4}+{0} = \color{#9D38BD}{4} $ If $\color{#9D38BD}{4}$ is divisible by $3$ , then ${40}$ must also be divisible by $3$ $\color{#9D38BD}{4}$ is not divisible by $3$, therefore ${661999}$ must not be divisible by $3$.